圆括号周围的Python间距

Question

我试图删除正在打印的内容的括号。

这是我的打印功能

print("The text contains", totalChars, "alphabetic characters of which", numberOfe, "(", percent_with_e, "%)", "are 'e'.")

它打印得像这样

The text contains 5 alphabetic characters of which 5 ( 100.0 %) are 'e'.

但我需要它像这样打印

The text contains 5 alphabetic characters, of which 5 (100.0%) are 'e'. 

唯一的区别似乎是括号周围的间距。我无法从一开始就将空间移除。

Answer 1

更简单的方法是使用.format()方法格式化字符串：

print("The text contains {} alphabetic characters, of which {} ({}%) are 'e'".format(totalChars, numberOfe, percent_with_e))

如果您想继续使用逗号，则需要sep关键字参数：

print("The text contains ", totalChars, " alphabetic characters of which ", numberOfe, " (", percent_with_e, "%) ", "are 'e'.", sep="")

Answer 2

如果使用str.format，则可以更好地控制inter参数间距（print使用默认的单个空格）：

print("The text contains {} alphabetic characters\
       of which {} ({}%) are 'e'.".format(totalChars, numberOfe, percent_with_e))

Answer 3

如果您无法使用Personal执行此操作，请不要将它们作为不同的参数提供（使用format的默认sep。）< / p>

即，将' '转换为percent_with_e并加入str：

+

或者print("The text contains", totalChars, "alphabetic characters of which", numberOfe, "(" + str(percent_with_e) + "%)", "are 'e'.") ：

format

Answer 4

这是打印格式的问题。将多个参数传递给print函数时，它会自动插入空格。如果要格式化字符串，最好的方法是使用＆＃39;％＆＃39;操作

percent = "(%d%%)" % percent_with_e
print("The text contains", totalChars, "alphabetic characters of which", numberOfe, percent, "are 'e'.")